Magical Triangle - Think Outside The Box!

Many people requested this one. It was asked to 9th grade students in India who had only learned geometry, so you are supposed to solve it without using trigonometry. I thank Anand Gautam for discovering the incredible solution!

Polish translation credit: Tłumaczenie: Przemyslav

You can also solve the problem with trigonometry. This post does so for the 2nd part of the question:

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[cristiluchian]

Presh : 1+1=2
Me : ok, i got that one

[obowonkonobo]

Me: *solves the problem looking at the thumbnail using trig*
You: "the challenge is to do it without trig"
Me: 😲

[chl]

I cannot believe that I’m watching this to entertain myself.

[blackhole]

These videos* make me question whether I really deserve to be in college.

[photocopi]

I've always found those problems so cool! Just rotating a triangle was the key, and I would never think about it until you showed us. Great job Presh!

[carcaperu]

"Nothing out of the box is good, trust me"
Pandora

[bluechicken]

I always love "Out Of The Box" thinking like this. Thanks again Presh!

[lifeofphyraprun]

I actually thought about rotation and solved the first part.I was too lazy for the second part.But the point I wanna make is that due to your videos, Presh, I was able to get better at geometry quite a bit, and also in Math problem solving in general.I wanna thank you for that.My creativity also actually increased.Please keep coming up with such good Math videos.

[ashwinidasanchal]

3:37 🤯🤯🤯🤯🤯 being an Indian student that amazed me

[fedryfirman.a]

Beautiful. The problem, the explanation, and everything this channel brings is pure amazing

[fabriciolopes]

There is another elegant way to do this question, this was the way i did this question when it first came to me:
My solution is about to prove that this EF segment defined by a 45° in the square is aways tangent to the quarter of circunference with unity radious and center in A.
To do this take the quarter of circunference and trace a tangent segment of it, now with the intersection point "P" trace AP and look that the pairs of triangles ADE, APE and ABF, APF are congruents by the Side Side Angle case (SSA - common sides, unity side and straight angle). With it, see the congruence between DAE, PAE and PAF, BAP angles, nevertheless how DAE+PAE+PAF+BAP=90° and DAE=PAE, PAF=BAP, then PAE+PAF=45°, proving that EAF will aways be 45°.
So, we can guarantee that if we pass a quarter circunference in the problem with radious 1 and center A it will tangent EF. Doing this, let the intersection be an X point, now observe the AXF and ABF triangles - they are congruent by the same case before - and see that AFB and AFX(AFX=AFE) angles are both 70°, letting angle AEF= 65°(AFE+FEA+EAF=180° - EAF=45°, AFE=70°). (:

[elyn]

i'm not find your channel early?... I just want to entertain myself with this something useful rather than a meme video or drama that wasting my time a lot while quarantine 😭😭😭

[charlesdang]

Nice puzzle. I was content to solve this using brute force and was surprised to find whole number answers. This of course meant there was a more elegant solution. Thanks for showing the reflection trick. Immediately this brings to mind folding tricks, origami as one. if we fold the two flaps on either side, they would fit neatly within the larger triangle in the middle. Super neat!

[avikdas]

Surprised that the solution doesn't involve the gougu theorem.

[ascaniuspotterhead]

I’d have chosen a brute force method and simply made equations over equations and then slowly eliminating the unknown parts till I can solve it. But your way is much faster and easier. Thank you for making this awesome content.

[xz]

Presh's way ("half angle" model) is good, I just want to show alternative, which is good too. Draw FH perpendicular to AE and meet @ H, thus A, B, F, H and E, C, F, H are co-circular (concyclic), respectively.

So ∠HBF=∠HAF=45° (circumferential angle), so BH must be the diagonal of ABCD; connect DH (which must be diagonal), so ∆ADH congruent to ∆CDH, thus ∠DAH=∠DCH=90-45-20=25°, hence, ∠HFE =25° too, bcz it's circumferential angle, so we get ∠AEF=90-25=65°

As for second part of the problem, we can use Gougu theorems and similar triangles (∆ADE and ∆FHE) to get answer, that's 2 (for you guys to do 😁).

[lilac]

As a ninth grader in India, I'll admit that I was unable to solve this one, but I was able to understand exactly how you reached the conclusions. I just didn't think about rotation. Thanks for sharing this problem, we have tons of questions like these in our book since almost half our chapters in the book are geometry this year (Quadrilaterals, Triangles, Lines and Angles, Circles, etc).

[thenegotiator]

I’ve been watching you channel for a long time that i am able now to every problem you upload. It feels amazing when you say did you figure it out and the answer is yes :)

[vaibhavgt]

Thinking “Out Of The Box” is something that I should apply in geometry too.
Never thought Congruency will be used.

Love from Haridwar (India)❤️

[pointless]

Also we can find the angle by folding ADE and ABF when we draw the height of AEF